Round away from zeroeig36

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Round away from zero

Inspired by Round towards zero.

Given a number input via any reasonable method, round the number "away from zero" - positive numbers round up, and negative numbers round down.

You can output a floating-point number with the decimal point (e.g. 42.0) if desired. (Or even have some test cases output floating-point and some output integer, if it makes your answer shorter.)

Standard loopholes are not allowed, etc etc.

Test cases

-99.9 => -100
-33.5 => -34
-7    => -7
-1.1  => -2
0     => 0
2.3   => 3
8     => 8
99.9  => 100
42.0  => 42
-39.0 => -39

Sandbox Link

code-golf number
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  • \\$\\begingroup\\$ if we're taking numbers in a string context, such as STDIN, do we need to support the .0 as the test cases seem to suggest? \\$\\endgroup\\$ – Jo King 3 hours ago

17 Answers 17

active oldest votes
5
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Jelly, 4 bytes

ĊṠ¡Ḟ

A monadic Link accepting a number which yields an integer.

Try it online! Or see a test-suite.

How?

ĊṠ¡Ḟ - Link: number, N
  ¡  - repeat...
 Ṡ   - ...number of times: sign of N (repeating -1 is the same as 0 times)
Ċ    - ...action: ceiling
   Ḟ - floor (that)
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  • \\$\\begingroup\\$ So how exactly does ¡ work for negative numbers? I don't think it's documented \\$\\endgroup\\$ – caird coinheringaahing 8 hours ago
  • 1
    \\$\\begingroup\\$ It's not documented on Jelly's wiki, but ¡s repetitive nature is implemented with a for index in range(repetitions) loop in the code. range([stop=]-1) is empty since start defaults to 0 and step defaults to 1 and "For a positive step, the contents of a range r are determined by the formula r[i] = start + step*i where i >= 0 and r[i] < stop." docs \\$\\endgroup\\$ – Jonathan Allan 7 hours ago
  • \\$\\begingroup\\$ ¡'s behavior relies on that of Python's range, and range(-1).__iter__().__next__() immediately throws StopIteration. \\$\\endgroup\\$ – Unrelated String 7 hours ago
4
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R, 32 bytes

x=scan()
sign(x)*ceiling(abs(x))
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  • \\$\\begingroup\\$ 31 bytes -- very nice answer! \\$\\endgroup\\$ – Giuseppe 8 hours ago
4
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Jelly, 5 bytes

Ṡ×AĊ$

Try it online!

This ports recursive's Stax answer into Jelly

Jelly, 6 bytes

ĊḞṠ‘$?

Try it online!

I feel like there is a shorter way than this, but it works, and isn't a port of another answer.

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  • \\$\\begingroup\\$ ĊḞ>?0 would work as your 6 does. \\$\\endgroup\\$ – Jonathan Allan 7 hours ago
  • 1
    \\$\\begingroup\\$ AĊ×Ṡ is 4 and functionally identical to your first answer. \\$\\endgroup\\$ – Nick Kennedy 7 hours ago
2
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JavaScript (ES6), 20 bytes

n=>n%1?n<0?~-n:-~n:n

Try it online!

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2
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Stax, 6 bytes

å├╪∙Bß

Run and debug it

Procedure:

  1. Absolute value
  2. Ceiling
  3. Multiply by original sign
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2
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Python 3, 24 bytes

lambda i:i-i%(1,-1)[i>0]

Try it online!

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2
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Retina 0.8.2, 38 bytes

\\.0+
.
\\b9+\\..
0$&
T`9d`d`.9*\\..
\\..*

Try it online! Link includes test cases. Explanation:

\\.0+
.

Delete zeroes after the decimal point, to ensure that the number is not an integer; the next two matches fail if there are no digits after the decimal point.

\\b9+\\..
0$&

If the integer part is all 9s, prefix a 0 to allow the increment to overflow.

T`9d`d`.9*\\..

Increment the integer part of the number.

\\..*

Delete the fractional part of the number.

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1
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Vim, 36 bytes/keystrokes

:s/-/-<Space>
:g/\\..*[1-9]/norm <C-v><C-a>lD
:s/<Space><cr>

Try it online! or Verify all Test Cases!

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1
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Runic Enchantments, 18 bytes

i:'|A:1+µ-'fA},*@

Try it online!

"Adds" (away from zero) 0.999999 and floors the result. µ is the closest thing to an infinitesimal in language's operators.

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  • 1
    \\$\\begingroup\\$ This outputs nothing for 0 \\$\\endgroup\\$ – Jo King 6 hours ago
  • 1
    \\$\\begingroup\\$ @JoKing Oof. Good catch. It's doing a divide by input to get the "sign" of the input value, which of course, divides by 0 when the input is 0. There's no (good) way around that right now. Will need this commit first. I'll poke Dennis (side benefit, the answer will get shorter). \\$\\endgroup\\$ – Draco18s 5 hours ago
1
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Perl 6, 18 bytes

{$_-.abs%-1*.sign}

Try it online!

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1
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Excel, 13 bytes

=ROUNDUP(A1,)

Alternative

=EVEN(A1*2)/2
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1
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J, 6 bytes

**>.@|

Try it online!

Just a 1 character change from my answer on the cousin question.

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0
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Brachylog, 7 bytes

⌋₁ℤ₁|⌉₁

Try it online!

or ⌉₁ℕ₁|⌋₁.

⌋₁         The input rounded down
  ℤ₁       is an integer less than -1
    |      and the output, or, the input
     ⌉₁    rounded up is the output.
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0
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Perl 6, 19 bytes

{$_+|0+.sign*?/\\./}

Try it online!

Not the shortest solution, but I'm working on it. Basically this truncates the number, then adds one away from zero if the number was not whole to begin with

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0
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Java (OpenJDK 8), 43 bytes

a->{return(int)((int)a/a<1?a>0?a+1:a-1:a);}

Try it online!

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  • 1
    \\$\\begingroup\\$ The lambda function can be written without using an explicit return statement. \\$\\endgroup\\$ – Joel 6 hours ago
  • \\$\\begingroup\\$ @Joel is indeed right. And you can save 4 additional bytes changing the (int)a/a<1 to a%1!=0: 30 bytes \\$\\endgroup\\$ – Kevin Cruijssen 52 secs ago
0
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Perl 5 -pF/\\./, 24 bytes

$_&&=$F[0]+$_*(@F-1)/abs

Try it online!

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0
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C, 94 bytes

#include<math.h>
main(){float a;scanf("%e",&a);printf("%f",(-2*signbit(a)+1)*ceil(fabs(a)));}

I compiled with

gcc a.c
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